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Prior Knowledge Check 1) Factorise each polynomial: a) x 2 6x + 5 b) x 2 16 c) 9x 2 25 2) Simplify the following algebraic fractions fully: a) x 2 9 x 2 +9x+18 b) 2x2 +5x 12 6x 2 7x 3 c) x2 x 30 x 2 +3x+18 (x 1)(x 5) (x + 4)(x 4) (3x 5)(3x + 5) x 3 x + 6 x + 4 3x + 1 x + 5 x + 3 3) For any integers n and m, decide whether the following will always be odd, always be even, or could be either: a) 8n b) n m c) 3m d) 2n 5 Even Either Either Odd

You need to be able to prove statements by contradiction To prove a statement by contradiction, you need to follow these steps: 1) Assume the statement is false 2) Use logical steps to show that this leads to an impossible outcome, or one that contradicts the original statement The statement that shows the original assumption is false is known as the negation of the statement Prove by contradiction that there is no greatest odd integer Assumption: There is a greatest odd integer, n We need to use logical steps to reach a contradiction/impossibility If n is an integer, then n + 2 is also an integer n + 2 > n n + 2 = odd + even = odd So n + 2 would be an odd integer, and is greater than n. Therefore, there is no greatest odd integer. 1A

You need to be able to prove statements by contradiction To prove a statement by contradiction, you need to follow these steps: 1) Assume the statement is false 2) Use logical steps to show that this leads to an impossible outcome, or one that contradicts the original statement The statement that shows the original assumption is false is known as the negation of the statement Prove by contradiction that if n 2 is even, then n must be even Assumption: There exists a number n such that n is odd, but n 2 is even. We need to use logical steps to reach a contradiction/impossibility If n is odd then it can be written in the form 2k + 1 Therefore n 2 = 2k + 1 2 = 4k 2 + 4k + 1 = 2(2k 2 + 2k) + 1 So n 2 is odd, which contradicts the original statement that if n is odd, then n 2 can be even 1A

You need to be able to prove statements by contradiction Prove by contradiction that 2 is an irrational number To prove a statement by contradiction, you need to follow these steps: 1) Assume the statement is false 2) Use logical steps to show that this leads to an impossible outcome, or one that contradicts the original statement The statement that shows the original assumption is false is known as the negation of the statement Assumption: 2 is a rational number, and can therefore be expressed as a, with the fraction is b its simplest form We need to use logical steps to reach a contradiction/impossibility 2 = a b 2 = a2 b 2 2b 2 = a 2 2b 2 = (2n) 2 Square both sides Multiply by b 2 Replace a with 2n Square This means that a 2 is even, 2bso 2 therefore = 4n 2 a must also be even Divide by 2 This means that b 2 is also even, so This means that a can bbe 2 = written 2n 2 as 2n, where n is a therefore b must also be even different integer 1A

You need to be able to prove statements by contradiction To prove a statement by contradiction, you need to follow these steps: 1) Assume the statement is false 2) Use logical steps to show that this leads to an impossible outcome, or one that contradicts the original statement The statement that shows the original assumption is false is known as the negation of the statement Prove by contradiction that 2 is an irrational number Assumption: 2 is a rational number, and can therefore be expressed as a, with the fraction is b its simplest form We have just shown that a and b are both even how does this contradict the original statement? If a and b are both even, then we can simplify the fraction a b However, our assumption was that 2 can be written as a, which has been fully simplified b What we showed contradicts this! Legend has it that Pythagoras believed that all roots can be written as rational numbers. A student proved otherwise, so he had the student killed! 1A

You need to be able to prove statements by contradiction To prove a statement by contradiction, you need to follow these steps: 1) Assume the statement is false 2) Use logical steps to show that this leads to an impossible outcome, or one that contradicts the original statement The statement that shows the original assumption is false is known as the negation of the statement Prove by contradiction that there are infinitely many Prime numbers Assumption: There is a finite number of Primes, with the largest being p n We need to use logical steps to reach a contradiction/impossibility Imagine we listed all the prime numbers: p 1, p 2, p 3,,, p n A number N will exist which is created by multiplying all the primes up to p n, and then adding 1 p 1 p 2 p 3,, p n This number will not be divisible by any of the primes, since 1 has been added Therefore, this number must either be Prime, or have a prime factor which was not on the original list Either way, the statement has been contradicted! 1A

You need to be able to multiply and divide Algebraic Fractions The rules for Algebraic versions are the same as for numerical versions When multiplying Fractions, you multiply the Numerators together, and the Denominators together a) b) c) Example Questions 1 3 2 5 a c b d 3 5 5 9 3 10 ac bd 15 45 It is possible to simplify a sum before you work it out. This will be vital on harder Algebraic questions 3 5 5 9 1 3 1 1 3 1 3 1B

You need to be able to multiply and divide Algebraic Fractions The rules for Algebraic versions are the same as for numerical versions When multiplying Fractions, you multiply the Numerators together, and the Denominators together It is possible to simplify a sum before you work it out. This will be vital on harder Algebraic questions Example Questions a c b a d) 1 1 c b x 1 3 e) 2 2 x 1 1 x 1 3 1 2 ( x1)( x1) 3 2( x 1) Factorise Multiply Numerator and Denominator 1B

You need to be able to multiply and divide Algebraic Fractions The rules for Algebraic versions are the same as for numerical versions When multiplying Fractions, you multiply the Numerators together, and the Denominators together It is possible to simplify a sum before you work it out. This will be vital on harder Algebraic questions a) Example Questions 5 1 6 3 5 3 6 1 15 6 5 2 When dividing Fractions, remember the rule, Leave, Change and Flip Leave the first Fraction, change the sign to multiply, and flip the second Fraction. 1B

You need to be able to multiply and divide Algebraic Fractions Example Questions The rules for Algebraic versions are the same as for numerical versions b) a b a c When multiplying Fractions, you multiply the Numerators together, and the Denominators together 1 a b 1 c a c b It is possible to simplify a sum before you work it out. This will be vital on harder Algebraic questions When dividing Fractions, remember the rule, Leave, Change and Flip Leave the first Fraction, change the sign to multiply, and flip the second Fraction. 1B

You need to be able to multiply and divide Algebraic Fractions Example Questions The rules for Algebraic versions are the same as for numerical versions When multiplying Fractions, you multiply the Numerators together, and the Denominators together It is possible to simplify a sum before you work it out. This will be vital on harder Algebraic questions When dividing Fractions, remember the rule, Leave, Change and Flip Leave the first Fraction, change the sign to multiply, and flip the second Fraction. c) 1 1 x2 3x6 x 2 4 x 16 x x4 3x6 2 2 x 16 x 2 ( x 4)( x 4) 1 x4 3( x2) 1 ( x 4) 3 Leave, Change and Flip Factorise Multiply the Numerators and Denominators 1B

You need to be able to add and subtract Algebraic Fractions The rules for Algebraic versions are the same as for numerical versions When adding and subtracting fractions, they must first have the same Denominator. After that, you just add/subtract the Numerators. c) Factorise so you can compare Denominators Multiply by (x - 1) Expand the bracket, and write as a single Fraction Simplify the Numerator Example Questions 3 4x x Factorise so 2 1 x 1 3 4x x 1 ( x 1)( x 1) 3( x 1) 4x ( x1) ( x 1) ( x1)( x 1) 3x34x ( x1)( x1) 7x 3 ( x1)( x1) you can compare Denominators Expand the bracket, and write as a single Fraction Simplify the Numerator 1C

You can split a fraction with two linear factors into Partial Fractions For example: x 1 (x + 3)(x + 1) 2 1 = when split up into Partial Fractions x + 3 x + 1 11 (x 3)(x + 2) A B = + when split up into Partial Fractions x 3 x + 2 You need to be able to calculate the values of A and B 1D

You can split a fraction with two linear factors into Partial Fractions Split 6x 2 (x 3)(x + 1) into Partial Fractions If x = -1: If x = 3: 6x 2 (x 3)(x + 1) A (x 3) A(x + 1) (x 3)(x + 1) = + + B (x + 1) B(x 3) (x 3)(x + 1) A x + 1 + B(x 3) (x 3)(x + 1) 6x 2 = A x + 1 + B(x 3) 8 = 4B 2 = B 16 = 4A Split the Fraction into its 2 linear parts, with numerators A and B Cross-multiply to make the denominators the same Group together as one fraction This has the same denominator as the initial fraction, so the numerators must be the same = 4 (x 3) 4 = A + 2 (x + 1) You now have the values of A and B and can write the answer as Partial Fractions 1D

You can also split fractions with more than 2 linear factors in the denominator Split 6x 2 + 5x 2 x(x 1)(2x + 1) into Partial fractions If x = 1 A(x 1)(2x + 1) x(x 1)(2x + 1) A x 6x 2 + 5x 2 x(x 1)(2x + 1) B + + x 1 C 2x + 1 B(x)(2x + 1) + x(x 1)(2x + 1) C(x)(x 1) + x(x 1)(2x + 1) A x 1 2x + 1 + B x 2x + 1 + C(x)(x 1) x(x 1)(2x + 1) 6x 2 + 5x 2 = 9 = 3B 3 = B Split the Fraction into its 3 linear parts A x 1 2x + 1 + B x 2x + 1 + C(x)(x 1) Cross Multiply to make the denominators equal Put the fractions together The numerators must be equal If x = 0 If x = -0.5 = 2 x 2 = 2 = A 3 = 4 = C 3 + x 1 A 0.75C 4 2x + 1 You can now fill in the numerators 1D

You need to be able to split a fraction that has repeated linear roots into a Partial Fraction For example: 3x 2 4x + 2 x + 1 (x 5) 2 = A (x + 1) + B (x 5) + C (x 5) 2 when split up into Partial Fractions The repeated root is included once fully and once broken down 1E

You need to be able to split a fraction that has repeated linear roots into a Partial Fraction Split 11x 2 + 14x + 5 (x + 1) 2 (2x + 1) into Partial fractions A (x + 1) A(x + 1)(2x + 1) (x + 1) 2 (2x + 1) = 11x 2 + 14x + 5 + 11x 2 + 14x + 5 (x + 1) 2 (2x + 1) B + + (x + 1) 2 B(2x + 1) (x + 1) 2 (2x + 1) C (2x + 1) + C(x + 1) 2 (x + 1) 2 (2x + 1) A x + 1 2x + 1 + B 2x + 1 + C(x + 1)2 (x + 1) 2 (2x + 1) = A x + 1 2x + 1 + B 2x + 1 + C(x + 1) 2 Split the fraction into its 3 parts Make the denominators equivalent Group up The numerators will be the same If x = -1 2 = B 2 = B At this point there is no way to cancel B and C to leave A by substituting a value in Choose any value for x (that hasn t been used yet), and use the values you know for B and C to leave A If x = -0.5 0.75 = 3 = C 0.25C If x = 0 5 = 1A + 1B + 1C = 4 (x + 1) 5 = A 2 + 3 4 = A 2 3 + (x + 1) 2 (2x + 1) Sub in the values of A, B and C 1E

If you have an improper fraction, it must first be converted into a mixed fraction before you can express it in partial fractions You can solve these kinds of problems by using algebraic long division Alternatively, you can use the relationship: Given that x3 +x 2 7 Ax 2 + Bx + C + D, x 3 x 3 find the values of A, B, C and D Using algebraic long division x 3 x 2 + 4x + 12 x 3 + x 2 + 0x 7 x 3 3x 2 4x 2 + 0x 7 4x 2 12x 12x 7 12x 36 29 F x = Q x divisor + remainder 37 = 12 3 + 1 x 3 + x 2 7 x 3 x 2 + 4x + 12 remainder 29 Write the remainder over the divisor (as you would if dividing with numbers) x 3 + x 2 7 x 3 x 2 + 4x + 12 + 29 x 3 1F

If you have an improper fraction, it must first be converted into a mixed fraction before you can express it in partial fractions Given that: x 4 + x 3 + x 10 Ax 2 + Bx + C x 2 + 2x 3 + Dx + E find the values of A, B, C, D and E. You can solve these kinds of problems by using algebraic long division Compare x 4 terms x 4 = Ax 4 A = 1 Compare x 3 terms x 3 = 2x 3 + Bx 3 B = 1 Alternatively, you can use the relationship: Updated relationship x 4 + x 3 + x 10 x 2 + x Bx + + 5 C Cx 2 x+ 2 + 2x 2x 3 3 + + 12x Dx Dx + E 5E E F x = Q x divisor + remainder Compare x 2 terms 0x 2 = Cx 2 3x 2 2x 2 C = 5 Compare x terms x = 3x + 10x + Dx D = 12 Compare constant 10 = 15 + E E = 5 1F

You can split an improper fraction into Partial Fractions. You will need to divide the numerator by the denominator first to find the whole part 22 35 = 1 5 + 3 7 A regular fraction being split into 2 components 57 20 = 1 2 + 4 + 3 5 A top heavy (improper) fraction will have a whole number part before the fractions 1G

You can split an improper fraction into Partial Fractions. You will need to divide the numerator by the denominator first to find the whole part Split 3x 2 3x 2 (x 1)(x 2) = x 2 3x + 2 3x 2 3x 2 3 3x 2 9x + 6 3x 2 3x 2 x 2 3x + 2 Divide the numerator by the denominator to find the whole part 3x 2 3x 2 (x 1)(x 2) into Partial fractions Remember, Algebraically an improper fraction is one where the degree (power) of the numerator is equal to or exceeds that of the denominator 6x 8 3x 2 3x 2 (x 1)(x 2) = 6x 8 3 + (x 1)(x 2) A (x 1) = 6x 8 = A x 2 + B(x 1) If x = 2 4 = B If x = 1 2 = A 2 = A + B (x 2) A x 2 + B(x 1) (x 1)(x 2) Now rewrite the original fraction with the whole part taken out Split the fraction into 2 parts (ignore the whole part for now) Make denominators equivalent and group up The numerators will be the same = 3 + 2 (x 1) + 4 (x 2) 1G